Please help me with these 2 questions!

8. For all pairs of real numbers P and Q where<br> P = 2Q + 9, Q = ?

Gnom [1K]

Answer:

Q = (P-9)/2

Step-by-step explanation:

5 0

2 years ago

The price p (in dollars) and the demand X for a particular clock radio are related by the equation X = 3000 -10p. Find and answe

alukav5142 [94]

The price p based on the equation given is p = 3000 - 0.1x.

<h3>How to express the price?</h3>

The equation given in the question is x = 3000 - 10p. The price(p) will be:

x = 3000 - 10p.

Make p the subject of the formula

x + 10p = 3000

10p = 3000 - x

p = (3000 - x)/10

p = 3000 - 0.1x

The revenue will be price multiplied by quantity. This will be:

= (3000 - 0.1x) × x

= 3000x - 0.1x²

The marginal revenue will be calculated after differentiating. This will be:

= 3000x - 0.1x²

= 3000 - 0.2x

The demand which is the quantity based on the information will be:

3000 - 0.2x = 0

0.2x = 3000

x = 1500

Learn more about demand on:

brainly.com/question/1245771

#SPJ1

6 0

1 year ago

Find all the solutions for the equation:

Contact [7]

$2y^2\,\mathrm dx-(x+y)^2\,\mathrm dy=0$

Divide both sides by $x^2\,\mathrm dx$ to get

$2\left(\dfrac yx\right)^2-\left(1+\dfrac yx\right)^2\dfrac{\mathrm dy}{\mathrm dx}=0$

$\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{2\left(\frac yx\right)^2}{\left(1+\frac yx\right)^2}$

Substitute $v(x)=\dfrac{y(x)}x$ , so that $\dfrac{\mathrm dv(x)}{\mathrm dx}=\dfrac{x\frac{\mathrm dy(x)}{\mathrm dx}-y(x)}{x^2}$ . Then

$x\dfrac{\mathrm dv}{\mathrm dx}+v=\dfrac{2v^2}{(1+v)^2}$

$x\dfrac{\mathrm dv}{\mathrm dx}=\dfrac{2v^2-v(1+v)^2}{(1+v)^2}$

$x\dfrac{\mathrm dv}{\mathrm dx}=-\dfrac{v(1+v^2)}{(1+v)^2}$

The remaining ODE is separable. Separating the variables gives

$\dfrac{(1+v)^2}{v(1+v^2)}\,\mathrm dv=-\dfrac{\mathrm dx}x$

Integrate both sides. On the left, split up the integrand into partial fractions.

$\dfrac{(1+v)^2}{v(1+v^2)}=\dfrac{v^2+2v+1}{v(v^2+1)}=\dfrac av+\dfrac{bv+c}{v^2+1}$

$\implies v^2+2v+1=a(v^2+1)+(bv+c)v$

$\implies v^2+2v+1=(a+b)v^2+cv+a$

$\implies a=1,b=0,c=2$

Then

$\displaystyle\int\frac{(1+v)^2}{v(1+v^2)}\,\mathrm dv=\int\left(\frac1v+\frac2{v^2+1}\right)\,\mathrm dv=\ln|v|+2\tan^{-1}v$

On the right, we have

$\displaystyle-\int\frac{\mathrm dx}x=-\ln|x|+C$

Solving for $v(x)$ explicitly is unlikely to succeed, so we leave the solution in implicit form,

$\ln|v(x)|+2\tan^{-1}v(x)=-\ln|x|+C$

and finally solve in terms of $y(x)$ by replacing $v(x)=\dfrac{y(x)}x$ :

$\ln\left|\frac{y(x)}x\right|+2\tan^{-1}\dfrac{y(x)}x=-\ln|x|+C$

$\ln|y(x)|-\ln|x|+2\tan^{-1}\dfrac{y(x)}x=-\ln|x|+C$

$\boxed{\ln|y(x)|+2\tan^{-1}\dfrac{y(x)}x=C}$

7 0

3 years ago

If the altitude of an equilateral triangle is 9 what is the perimeter

Volgvan

The angles in an equilateral triangle are 60 degrees.

sin 60 = 9 / h where h is hypotenuse ( = side of the triangle)

h = 9 / sin 60 = 10.392

Perimeter = 3 * h = 31.18 to nearest hundredth

3 0

3 years ago

Read 2 more answers

WILL MARK U AS BRAINLIEST PLZ HELP

Radda [10]

Answer:

112 if you don't include the triangle sides as they are not necessary, but 124 if you include one triangle and 136 if you include both triangle sides

5 0

3 years ago

Please help me with these 2 questions! ​

Please help me with these 2 questions!